# 1.9: Pigeonhole Principle

- Page ID
- 17370

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INVESTIGATE!!

- Suppose there are
*n*people at a party, with*n*at least 2. Show that there are two people that have the same number of friends. - Suppose 5 points are selected from inside a \(1\times 1\) square. Prove that two of the points must be within \(\frac{1}{\sqrt{2}}\) of each other.

Pigeonhole principle

- Simple version: If
*n+1*pigeons are placed in*n*pigeonholes, then at least one pigeonhole contains two or more pigeons. - General version: If
*n*or more pigeons are placed in*k*pigeonholes, then at least one pigeonhole contains \(\lceil\frac{n}{k}\rceil\) or more pigeons.

As a consequence, we can say that there must be two students at Notre Dame whose phone numbers end with the same four digits. Similarly, there must be two non-bald people in Chicago with the exact same number of hairs on their heads.

Example \(\PageIndex{1}\):

Seven distinct numbers are selected from the set \(\{1,2,3,\ldots,11\}\). Prove that two of these numbers must sum to 12.

Example \(\PageIndex{2}\):

Prove that among any 7 integers there must be two whose difference is divisible by 6.